Dilute sulfuric acid is electrolysed using inert electrodes

CHEMISTRY · Single correct · AP · Chemistry · Thermodynamics and Electrochemistry · Thermodynamics and Electrochemistry

Problem

Dilute sulfuric acid is electrolysed using inert electrodes. What are the ionic half-equations for the reactions that take place at each electrode?
positive electrodenegative electrode
A2H++2e−→H22\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_24OH−→2H2O+O2+4e−4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-
B2H++2e−→H22\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_24OH−+4H+→4H2O4\text{OH}^- + 4\text{H}^+ \rightarrow 4\text{H}_2\text{O}
C4OH−→2H2O+O2+4e−4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-2H++2e−→H22\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
D4OH−+4H+→4H2O4\text{OH}^- + 4\text{H}^+ \rightarrow 4\text{H}_2\text{O}2H++2e−→H22\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
(A) Positive: 2H++2e−→H22\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2; Negative: 4OH−→2H2O+O2+4e−4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^- (B) Positive: 2H++2e−→H22\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2; Negative: 4OH−+4H+→4H2O4\text{OH}^- + 4\text{H}^+ \rightarrow 4\text{H}_2\text{O} (C) Positive: 4OH−→2H2O+O2+4e−4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-; Negative: 2H++2e−→H22\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 (D) Positive: 4OH−+4H+→4H2O4\text{OH}^- + 4\text{H}^+ \rightarrow 4\text{H}_2\text{O}; Negative: 2H++2e−→H22\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2

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